Lecture Notes on Probability Theory and Random Processes by Jean Walrand

Author:Jean Walrand [Walrand, Jean]
Language: eng
Format: epub
Published: 2011-02-19T05:00:00+00:00

8.6. SOLVED PROBLEMS

137

The solution is a direct application of the definitions plus some algebra.

a. By definition,

M LE[X | Y = y] = arg maxxP [Y = y | X = x] = arg maxxP (x, y).

Hence, M LE[X | Y = 0] = 0 because P [Y = 0 | X = 0] = 0.7 > P [Y = 0 | X = 1] = 0.4.

Similarly, M LE[X | Y = 1] = 1 because P [Y = 1 | X = 1] = 0.6 > P [Y = 1 | X = 0] = 0.3.

Consequently,

M LE[X | Y ] = Y.

b. We know that

M AP [X | Y = y] = arg maxxP (X = x)P [Y = y | X = x] = arg maxxP (X = x)P (x, y).

Therefore,

M AP [X | Y = 0]

= arg maxx{g(x = 0) = P (X = 0)P (0, 0) = 0.7(1 − p), g(x = 1) = P (X = 1)P (1, 0) = 0.4p}

and

M AP [X | Y = 1]

= arg maxx{h(x = 0) = P (X = 0)P (0, 1) = 0.3(1 − p), h(x = 1) = P (X = 1)P (1, 1) = 0.6p}.

Consequently,







 0, if y = 0 and p < 7/11;





 1, if y = 0 and p ≥ 7/11;

M AP [X|Y = y] = 



 0, if y = 1 and p < 1/3;





 1, if y = 1 and p ≥ 1/3.

c. We know that









1, if L(y) = P (1, y)/P (0, y) > λ;

ˆ

X =

1 w.p. γ, if L(y) = P (1, y)/P (0, y) = λ;







0, if L(y) = P (1, y)/P (0, y) < λ.

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